$h(t) = 6t^{2}+7t-2$ $g(n) = -3n^{2}+3n-3(h(n))$ $ h(g(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-1)$ . Then we'll know what to plug into the outer function. $g(-1) = -3(-1)^{2}+(3)(-1)-3(h(-1))$ To solve for the value of $g$ , we need to solve for the value of $h(-1)$ $h(-1) = 6(-1)^{2}+(7)(-1)-2$ $h(-1) = -3$ That means $g(-1) = -3(-1)^{2}+(3)(-1)+(-3)(-3)$ $g(-1) = 3$ Now we know that $g(-1) = 3$ . Let's solve for $h(g(-1))$ , which is $h(3)$ $h(3) = 6(3^{2})+(7)(3)-2$ $h(3) = 73$